Question

Solve the system of linear equations using the substitution method. \(x+2 y-z=3\) \(2 x+4 y-2 z=6\) \(-x-2 y+z=-6\)

Short answer

The system of equations has no solution because the simplified forms of the equations lead to inconsistent results. One equation is always true, while the other is never true.

Step-by-step solution

Rearrange Equation 1

Rearrange first equation \(x+2 y-z=3\) to solve for \(x\), we get \(x=3-2y+z\). We will use this equation to substitute the value of \(x\) in the other two equations.

Substitute into Equation 2 and Equation 3

Substituting \(x=3-2y+z\) from the first equation into the second equation \(2x+4y-2z=6\) we get \(2(3-2y+z)+4y-2z=6\) which simplifies to \(6-4y+2z+4y-2z=6\), and in turn simplifies further to 6=6. Similarly, substituting \(x=3-2y+z\) into the third equation \(-x-2 y+z=-6\) we get \(-(3-2y+z)-2y+z=-6\) which simplifies to \(-3+2y-z-2y+z=-6\) and in turn simplifies further to -3=-6.

Evaluate the Result

After simplifying the second and third equations we can see there is an inconsistency. The equation 6=6 from the second equation is always true. But, the equation -3=-6 from the third equation is never true. So, this system of equations has no solution.

Key concepts

Substitution Method

The substitution method is a powerful tool in algebra used to find the values of variables in a system of linear equations. It involves solving one of the equations for one variable, and then substituting this expression into the other equations. By doing so, you reduce the number of equations by one, making the system easier to solve.

For instance, if you start with the equation \(x + 2y - z = 3\), you can solve for \(x\) to get \(x = 3 - 2y + z\). This new expression for \(x\) can replace all instances of \(x\) in the other equations of the system, effectively eliminating the variable \(x\) from those equations.

The main advantage of the substitution method is that it simplifies the system of equations, especially when you are dealing with coefficients of 1 or -1. However, it can become complicated with fractions or higher coefficients, where other methods like elimination might be more efficient. Remember, algebraic manipulation is key in utilizing the substitution method effectively, often requiring additional steps of simplification after substituting.

Inconsistent System

An inconsistent system of equations is one in which there are no solutions that satisfy all equations simultaneously. This usually means that the equations represent parallel lines (in two variables) or planes (in three or more variables) that never intersect. If you attempt to solve an inconsistent system, you will usually end up with a statement that is a contradiction, such as \(-3 = -6\), indicating that there is no solution.

In the given exercise, after substituting the value of \(x\) found from the first equation into the second and third equations, you reach contradictory statements. The second equation simplifies to a tautology, \(6 = 6\), which is true for all values. Yet, the third equation simplifies to the false statement \(-3 = -6\). This indicates that the given system of equations is inconsistent, thus having no solution where all three equations are satisfied at once. Recognizing an inconsistent system is important as it saves time by stopping further unnecessary calculations.

Algebraic Manipulation

Algebraic manipulation refers to the use of various algebraic techniques to rearrange, simplify, or solve equations. It's a fundamental skill in mathematics that involves operations such as distributing, combining like terms, and isolating variables. In the context of solving systems of linear equations, this skill becomes essential.

When you solve the given system using substitution, algebraic manipulation is used at every step. For example, after substituting \(x = 3 - 2y + z\) into the second equation, you distribute the 2 and then combine like terms to simplify the equation. Mastering these algebraic techniques enables you to clearly see the structure of equations and can lead you to identify inconsistencies or solutions more quickly.

Good algebraic manipulation can often make the difference between finding a solution or identifying that there is no solution, as seen in the aforementioned exercise. Accuracy in this manipulation also prevents errors that could lead to misinterpreting the system as consistent or inconsistent incorrectly.